有没有办法从servlet返回一个列表到html页面

Question

我想从html页面发送请求到servlet然后在java servlet中创建一个列表然后我想把这个列表返回到同一个html页面。

Answer 1

的问题是，你永远不会初始化HttpSession s变量：

@WebServlet(urlPatterns = {"/ShowPersonServlet"}) 
public class ShowPersonServlet extends HttpServlet { 

    HttpSession s; //null by default 
    //... 

    protected void processRequest(...) { 
     //... 
     //since never initialized, s is null 
     user.add((Person) s.getAttribute("person")); 
    } 

    @Override 
    protected void doGet(...) { 
     //... 
     //since never initialized, s is null 
     s.setAttribute("person",person); 
    } 
} 

化妆它工作解决方案：设置s

s = request.getSession(); 

现实世界解决方案的价值：

• 删除 Servlet中的所有字段，从不尝试处理servlet中的状态，除非它们是由容器（如EJB）管理的资源。
• 将您的HttpSession s的范围更改为每个方法的本地范围。另外，将其名称从s更改为session或更有用的东西。
• 将您的HTML代码移动到处理查看详细信息（如JSP）的组件，然后执行转发视图。

所以，你的代码应该是这样的：

@WebServlet(urlPatterns = {"/ShowPersonServlet"}) 
public class ShowPersonServlet extends HttpServlet { 
    /* 
    HttpSession s ; //moved as local variable 
    Person person = new Person(); //moved as local variable 
    private List<Person> user = new ArrayList<Person>(); //not sure where you want to store this 
    */ 
    protected void processRequest(HttpServletRequest request, HttpServletResponse response) 
     throws ServletException, IOException { 
     response.setContentType("text/html;charset=UTF-8"); 
     HttpSession session = request.getSession(); 
     List<Person> personList = (List<Person>) session.getAttribute("personList"); 
     if (personList == null) { 
      personList = new ArrayList<>(); 
      session.setAttribute("personList", personList); 
     } 
     personList.add((Person) session.getAttribute("person")); 
     /* 
     try (PrintWriter out = response.getWriter()) { 
      //removed to shorten this answer 
     } 
     */ 
     request.getRequestDispatcher("/showPerson.jsp").forward(request, response); 
    } 

    @Override 
    protected void doGet(HttpServletRequest request, HttpServletResponse response) 
     throws ServletException, IOException { 
     Person person = new Person(); 
     person.setKey(request.getParameter("txt_Key")); 
     person.setFirstName(request.getParameter("txt_firstName")); 
     person.setLastName(request.getParameter("txt_lastName")); 
     processRequest(request, response); 
     HttpSession session = request.getSession(); 
     session.setAttribute("person",person); 
    } 

    @Override 
    protected void doPost(HttpServletRequest request, HttpServletResponse response) 
     throws ServletException, IOException { 
     processRequest(request, response); 
    } 
} 

更多信息：

• How do servlets work? Instantiation, sessions, shared variables and multithreading
• StackOverflow Servlets wiki

Answer 2

你需要得到来自会话request，你c annot只是期望得到它与HttpSession s ; - 有s是null。

// Something like this (in processRequest), although I 
// would prefer a local session variable. 
s = request.getSession(); 

Answer 3

您需要提供异常的堆栈跟踪。但它看起来可能在两个地方：

1. s.setAttribute（“person”，person）; s从未设置;
2. out.println（“”+ p.getKey（）+“”）;其中属性“人”从未设置;

祝你好运。

Answer 4

var s的值未定义。它是空的，所以使用s.getAttribute("str")将会抛出异常。

HttpSession s ;

s = request.getSession(false);